Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
T gcd(T a, T b) {
  return b == 0 ? a : gcd(b, a % b);
}
const int MAXN = 1e5 + 1e4;
int n, pos_ = 0;
vector<int> g[MAXN];
vector<int> a[MAXN];
vector<int> ans;
vector<vector<int>> t(4 * MAXN);
vector<int> parent(MAXN), depth(MAXN), heavy(MAXN, -1), pos(MAXN), head(MAXN);
int dfs(int v) {
  int sz = 1;
  int h_sz = 0;
  for (int u : g[v]) {
    if (parent[v] == u) {
      continue;
    }
    parent[u] = v;
    depth[u] = depth[v] + 1;
    int cur_sz = dfs(u);
    sz += cur_sz;
    if (h_sz < cur_sz) {
      h_sz = cur_sz;
      heavy[v] = u;
    }
  }
  return sz;
}
void hld(int v, int root) {
  head[v] = root;
  pos[v] = pos_++;
  ;
  if (heavy[v] != -1) {
    hld(heavy[v], root);
  }
  for (int u : g[v]) {
    if (u == heavy[v] || u == parent[v]) {
      continue;
    }
    hld(u, u);
  }
}
void build(int v = 1, int l = 0, int r = n) {
  if (r - l == 1) {
    t[v] = a[l];
    while (((int32_t)(t[v]).size()) > 10) {
      t[v].pop_back();
    };
    return;
  }
  int m = (l + r) / 2;
  build(v * 2, l, m);
  build(v * 2 + 1, m, r);
  t[v].resize(20, INT_MAX);
  merge((t[v * 2]).begin(), (t[v * 2]).end(), (t[v * 2 + 1]).begin(),
        (t[v * 2 + 1]).end(), t[v].begin());
  while (((int32_t)(t[v]).size()) > 10 ||
         t[v][((int32_t)(t[v]).size()) - 1] == INT_MAX) {
    t[v].pop_back();
  }
}
void get_ans(int ql, int qr, int v = 1, int l = 0, int r = n) {
  if (ql <= l && r <= qr) {
    vector<int> ans_new(20, INT_MAX);
    merge((ans).begin(), (ans).end(), (t[v]).begin(), (t[v]).end(),
          ans_new.begin());
    for (int i = 0; i < 20; i++) {
      ;
    }
    while (((int32_t)(ans_new).size()) > 10 ||
           ans_new[((int32_t)(ans_new).size()) - 1] == INT_MAX) {
      ans_new.pop_back();
    }
    ans = ans_new;
    return;
  }
  if (qr <= l || r <= ql) {
    return;
  }
  int m = (l + r) / 2;
  get_ans(ql, qr, v * 2, l, m);
  get_ans(ql, qr, v * 2 + 1, m, r);
}
void query(int a, int b) {
  while (head[a] != head[b]) {
    if (depth[head[a]] > depth[head[b]]) {
      swap(a, b);
    };
    get_ans(pos[head[b]], pos[b] + 1);
    b = parent[head[b]];
  }
  if (depth[a] > depth[b]) {
    swap(a, b);
  };
  get_ans(pos[a], pos[b] + 1);
}
int32_t main() {
  {
    cin.tie(0);
    cout.tie(0);
    ios_base::sync_with_stdio(0);
  };
  int m, q;
  cin >> n >> m >> q;
  for (int i = 1; i < n; i++) {
    int v, u;
    cin >> v >> u;
    v--;
    u--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(0);
  hld(0, 0);
  for (int i = 0; i < m; i++) {
    int tmp;
    cin >> tmp;
    ;
    a[pos[tmp - 1]].push_back(i + 1);
  }
  build();
  while (q--) {
    int v, u, a;
    cin >> v >> u >> a;
    v--;
    u--;
    ans.clear();
    query(v, u);
    ;
    int k = min(a, ((int32_t)(ans).size()));
    cout << k;
    for (int i = 0; i < k; i++) {
      cout << " " << ans[i];
    }
    cout << "\n";
  }
  return 0;
}